Problems

Solution:

1. Notice that \(P = \begin{pmatrix} u \cos \alpha t\\ u \sin \alpha t - \frac12 g t^2 \end{pmatrix}\), so \begin{align*} && |OP|^2 &= u^2 \cos^2 \alpha t^2 + \left (u \sin \alpha t - \frac12 g t^2 \right)^2 \\ &&&= u^2 \cos^2 \alpha t^2 +u^2 \sin^2 \alpha t^2 - u \sin \alpha g t^3 +\frac14 g^2 t^4 \\ &&&= u^2 t^2 -u\sin \alpha g t^3 + \frac14g^2t^4 \\ && \frac{\d |OP|^2}{\d t} &= 2u^2 t - 3u \sin \alpha g t^2+g^2 t^3 \\ &&&= t \left (2u^2 - 3u \sin \alpha (gt)+(gt)^2\right) \\ && \Delta &= 9u^2 \sin^2 \alpha -4 \cdot 2u^2 \cdot 1 \\ &&&= u^2 (9\sin^2 \alpha -8) \\ \end{align*} Therefore if \(\sin \alpha < \frac{2\sqrt{2}}3\) the discriminant is negative, the quadratic factor is always positive and the distance \(|OP|\) is always increasing. Similarly, if \(\sin \alpha > \frac{2 \sqrt{2}}3\) then the derivative has a root. This means somewhere on its (possibly extended) trajectory \(OP\) is decreasing. This must be before it lands, since if it were after it 'landed' then both the \(x\) and \(y\) distances are increasing, therefore it cannot occur after it 'lands'.
2. Note that \(Q = \begin{pmatrix} -v t \\0 \end{pmatrix}\) \begin{align*} && |QP|^2 &= (u \cos \alpha t+vt)^2 + \left (u \sin \alpha t - \frac12 g t^2 \right)^2 \\ &&&= u^2 \cos^2 \alpha t^2+2u\cos \alpha v t^2 + v^2 t^2 +u^2 \sin^2 \alpha t^2 - u \sin \alpha g t^3 +\frac14 g^2 t^4 \\ &&&= (u^2+2u v \cos \alpha+v^2) t^2 - u \sin \alpha g t^3 + \frac14 g^2 t^4 \\ \\ \Rightarrow && \frac{\d |QP|^2}{\d t} &= 2(u^2+u v \cos \alpha+v^2) t - 3u \sin \alpha g t^2 + g^2 t^3 \\ &&&= t \left ( 2(u^2+2u v \cos \alpha+v^2) - 3u \sin \alpha (g t) + (g t)^2\right) \\ && \Delta &= 9u^2 \sin^2 \alpha - 8(u^2+2u v \cos \alpha+v^2) \\ &&&= (9 \sin^2 \alpha -8)u^2 - 16v \cos \alpha u - 8v^2 \\ &&&= \left (( \sin \alpha-2\sqrt{2}\cos \alpha)u-2\sqrt{2} v \right) \left ( ( \sin \alpha+2\sqrt{2}\cos \alpha)u+2\sqrt{2} v \right) \end{align*} Since the second bracket is clearly positive, the first bracket must be negative (for \(\Delta < 0\) and our derivative to be positive), ie \(2\sqrt{2} v > ( \sin \alpha-2\sqrt{2}\cos \alpha)u\)

Solution: First note that \((\sqrt{6} - \sqrt{2})^2 = 8 - 2\sqrt{12} = 8 - 4\sqrt{3}\) and since \(49 > 16 \times 3\) \(\sqrt{6}-\sqrt{2} > 1\). Therefore we can assume without loss of generality that \(P\) and \(Q\) both do not lie on the same side as each other, a side containing \(O\), otherwise one of those lengths would be \(1 \text{ cm} < (\sqrt{6}-\sqrt{2}) \text{ cm} \). Let \(O = (0,0)\), \(P = (1,x)\), \(Q = (y,1)\), then our lengths squared are: \(1 + x^2, 1 + y^2, (1-x)^2+(1-y)^2\). To maximise the length of the smallest side, each side should be equal in length (otherwise we could increase the length of the smallest side by moving the point between the shortest side and the longest side (without affecting the other side). Therefore \(x = y\) and \(1+x^2 = 2(1-x)^2 \Rightarrow x^2-4x+1 = 0 \Rightarrow x = 2 - \sqrt{3} \). Therefore the distances are all \(\sqrt{1+7-4\sqrt{3}} = \sqrt{8-4\sqrt{3}} = (\sqrt{6}-\sqrt{2}) \text{ cm}\)

Solution: The plane is the same as the plane \((\mathbf{r} - \mathbf{a}) \cdot \mathbf{a} = 0\), ie the plane through \(\mathbf{a}\) whose normal is parallel to \(\mathbf{a}\) The distance from \(\mathbf{r}\) to the plane therefore is \(\lambda\) where \(\mathbf{r}+\lambda \frac{1}{a}\mathbf{a}\) must be on the plane, ie \((\mathbf{r}+\frac{\lambda}{a} \mathbf{a} - \mathbf{a})\cdot \mathbf{a} = 0 \Rightarrow \lambda = \frac{a^2-\mathbf{a} \cdot \mathbf{r}}{a}\) But if \(\mathbf{a} \cdot \mathbf{r} = a^2 - ar\) then \(\lambda = r\), ie the distance to the plane is the same as the distance to the origin. \(\mathbf{q} = k \mathbf{p}\) and so \(\mathbf{a} \cdot k \mathbf{p} + a |k|p = a^2\) if \(k > 0\) we will find \(k = 1\) the position vector we already know about, therefore suppose \(k < 0\) so: \begin{align*} && \mathbf{a} \cdot k \mathbf{p} - ka p &= a^2 \\ \Rightarrow && k(a^2-ap)-kap &= a^2 \\ \Rightarrow && k(a^2-2ap) &= a^2 \\ \Rightarrow && k &= \frac{a^2}{a^2-2ap} \end{align*} Therefore \(\mathbf{q} = \frac{a^2}{a^2-2ap} \mathbf{p}\) The line through \(O\) orthogonal to \(\mathbf{p}\) and \(\mathbf{a}\) will be parallel to \(\mathbf{a} \times \mathbf{p}\). Therefore we should consider points of the from \(s \mathbf{a} \times \mathbf{p}\) on the surface \(S\). \begin{align*} && s\mathbf{a} \cdot ( \mathbf{a} \times \mathbf{p}) + sa^2p |\sin \theta| &= a^2 \end{align*} The angle between \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{p}}{ap} = \frac{a^2-ap}{ap} \Rightarrow |\sin \theta| = \sqrt{1-\frac{(a-p)^2}{p^2}} = \frac{1}{p} \sqrt{2ap-a^2}\) Therefore \(sa^2 \sqrt{2ap-a^2} = a^2 \Rightarrow s = \frac{1}{\sqrt{2ap-a^2}}\) and so the points are as required. Noting that \(|\mathbf{p} \times \mathbf{t}| = |\frac{1}{p \sin \theta}\mathbf{p} \times (\mathbf{p} \times \mathbf{a}) | = |\frac{1}{p \sin \theta}p^2a \sin \theta | = pa\) The area of triangle \(PQT\) is : \begin{align*} \frac12 | (\mathbf{p} - \mathbf{t}) \times (\mathbf{q} - \mathbf{t}) | &= \frac12 |\mathbf{p} \times \mathbf{q} - \mathbf{t} \times \mathbf{q} - \mathbf{p} \times \mathbf{t} - \mathbf{t} \times \mathbf{t}| \\ &= \frac12 |\mathbf{t} \times (\mathbf{p} - \mathbf{q})| \\ &= \frac12 \cdot (1 - \frac{a^2}{a^2-2ap})| \mathbf{t} \times \mathbf{p}| \\ &= \frac12 \frac{2ap}{a^2-2ap} \cdot ap \\ &= \frac{ap^2}{a^2-ap} \end{align*}

Solution: Consider the angle from the origin, then \(P = (\cos \theta, \sin \theta)\) where \(\theta \sim U(0, 2\pi)\), and \(X = \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta}\) \begin{align*} \mathbb{E}[X] &= \int_0^{2\pi} \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta} \frac1{2\pi} \d \theta \\ &= \frac1{2\pi}\int_0^{2\pi} \sqrt{2 - 2\cos \theta} \d \theta \\ &= \frac{1}{2\pi}\int_0^{2\pi} \sqrt{4\sin^2 \frac{\theta}{2}} \d \theta \\ &= \frac{1}{\pi}\int_0^{2\pi} \left |\sin \frac{\theta}{2} \right| \d \theta \\ &= \frac{1}{\pi} \left [ -2\cos \frac{\theta}{2} \right]_0^{2\pi} \\ &= \frac1{\pi} \l 2 + 2\r \\ &= \frac{4}{\pi} \end{align*} \begin{align*} \mathbb{E}(X^2) &= \frac1{2\pi}\int_0^{2\pi} (\cos \theta - 1)^2 + \sin^2 \theta \d \theta \\ &= \frac1{2\pi}\int_0^{2\pi} 2 - 2 \cos \theta \d \theta \\ &= \frac{4\pi}{2\pi} \\ &= 2 \\ \end{align*} \(\Rightarrow\) \(\mathrm{Var}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2 = 2 - \frac{16}{\pi^2} = \frac{2\pi^2 - 16}{\pi^2}\).

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